Problem 23 - Calculus: Early Transcendentals

Question

Evaluate the integral. \( \displaystyle \int^0_{-2} \biggl( \frac{1}{2}t^4 + \frac{1}{4}t^3 - t \biggr) \,dt \)

Step-by-Step Solution

Answer

The value of the integral is 4.2.

Step 1: Identify the Integral

We need to evaluate the definite integral \( \int^0_{-2} \left( \frac{1}{2}t^4 + \frac{1}{4}t^3 - t \right) \ dt \). This involves finding the antiderivative of the given polynomial expression and then evaluating it at the upper and lower limits of the integral.

Step 2: Antidifferentiate the Polynomial

To find the antiderivative of the polynomial, we apply the power rule for each term separately:1. For \( \frac{1}{2} t^4 \), the antiderivative is \( \frac{1}{2} \cdot \frac{t^5}{5} = \frac{1}{10} t^5 \).2. For \( \frac{1}{4} t^3 \), the antiderivative is \( \frac{1}{4} \cdot \frac{t^4}{4} = \frac{1}{16} t^4 \).3. For \( -t \), the antiderivative is \( -\frac{t^2}{2} \).Thus, the antiderivative of the function is \( F(t) = \frac{1}{10} t^5 + \frac{1}{16} t^4 - \frac{1}{2} t^2 \).

Step 3: Evaluate the Antiderivative at the Bounds

Now we evaluate the antiderivative \( F(t) = \frac{1}{10} t^5 + \frac{1}{16} t^4 - \frac{1}{2} t^2 \) at the bounds \( t = 0 \) and \( t = -2 \):- At \( t = 0 \): \[ F(0) = \frac{1}{10}(0)^5 + \frac{1}{16}(0)^4 - \frac{1}{2}(0)^2 = 0 \]- At \( t = -2 \): \[ F(-2) = \frac{1}{10}(-2)^5 + \frac{1}{16}(-2)^4 - \frac{1}{2}(-2)^2 \] Simplifying further: \[ F(-2) = \frac{1}{10}(-32) + \frac{1}{16}(16) - \frac{1}{2}(4) \] \[ F(-2) = -\frac{32}{10} + 1 - 2 \] \[ F(-2) = -3.2 + 1 - 2 = -4.2 \]

Step 4: Apply the Fundamental Theorem of Calculus

Using the Fundamental Theorem of Calculus, the value of the integral is \( F(0) - F(-2) \):\[ \int_{-2}^{0} \left( \frac{1}{2}t^4 + \frac{1}{4}t^3 - t \right) \ dt = F(0) - F(-2) = 0 - (-4.2) = 4.2 \]

Key Concepts

Antiderivative

The concept of the antiderivative is fundamental in calculus, and it refers to finding a function whose derivative is the given function. It is also known as the "indefinite integral." In the context of evaluating a definite integral, we first identify the antiderivative of the given expression. For polynomial functions, finding the antiderivative is a straightforward process that involves reversing the differentiation rules.

• To find the antiderivative of a term like \( t^n \), consider the power rule for integration: Increase the power by one and divide by the new power.
• For example, the antiderivative of \( t^4 \) is \( \frac{t^5}{5} \), and thus \( \frac{1}{2}t^4 \) becomes \( \frac{1}{10}t^5 \) after considering the constant multiplication.

In our exercise, the polynomial expression \( \frac{1}{2}t^4 + \frac{1}{4}t^3 - t \) is antidifferentiated term by term resulting in \( \frac{1}{10}t^5 + \frac{1}{16}t^4 - \frac{1}{2}t^2 \), and this expression becomes crucial in evaluating the definite integral.

Power Rule

The power rule is a basic yet powerful tool in calculus, primarily when dealing with polynomial functions. It provides a simple formula for finding derivatives and antiderivatives of power functions.

• For differentiation, the power rule states: \( \frac{d}{dt}[t^n] = n t^{n-1} \). You simply bring down the power as a coefficient and reduce the power by one.
• For integration, which is essentially the reverse process, you add one to the exponent and divide by the new exponent: \( \int t^n \, dt = \frac{t^{n+1}}{n+1} + C \), where \( C \) is the constant of integration.

In our example, when finding the antiderivative of each term in the polynomial expression, applying the power rule is crucial. For instance, when antidifferentiating \( \frac{1}{4}t^3 \), the power increases from 3 to 4, and the coefficient is divided by the new power, leading to the term \( \frac{1}{16}t^4 \). This systematic approach simplifies finding antiderivatives.

Fundamental Theorem of Calculus

The Fundamental Theorem of Calculus elegantly connects the concept of differentiation and integration, forming a cornerstone of calculus. It states that if a function \( f \) is continuous over \([a, b]\), and \( F \) is an antiderivative of \( f \), then the definite integral of \( f \) from \( a \) to \( b \) is given by \( F(b) - F(a) \).

This theorem not only simplifies evaluating definite integrals but also provides insight into accumulation of quantities:

• First, you find an antiderivative \( F(t) \) of the function \( f(t) \).
• Then, you evaluate \( F \) at the bounds: calculate \( F(b) \) and \( F(a) \).
• The difference \( F(b) - F(a) \) gives the net area under the curve from \( a \) to \( b \).

In our exercise, once we have the antiderivative \( F(t) = \frac{1}{10}t^5 + \frac{1}{16}t^4 - \frac{1}{2}t^2 \), we apply the theorem: compute \( F(0) \) and \( F(-2) \) and subtract. This yields a result of 4.2, representing the signed area between the curve and the t-axis over the specified interval.